Solution (source code)

= Solution

Define $k_0=\omega/c_0$ and use the <outgoing acoustic square-root branch>
$$
\gamma^2=k^2-k_0^2,\qquad \operatorname{Re}\gamma>0\quad\text{for }\operatorname{Im}\omega<0.
$$
For positive real frequency reached from below, $\gamma=i\sqrt{k_0^2-k^2}$ on the propagating interval and is positive real for $|k|>k_0$. The outgoing field in the lower fluid has <pressure> amplitude $B e^{\gamma y}$, since $y<0$. If $\eta=\widehat\eta e^{i\omega t-ikx}$, the shared <normal velocity> is $i\omega\widehat\eta$. The <linear homentropic acoustic equations> therefore give
$$
i\omega\widehat\eta=-\frac{\gamma B}{i\omega\rho_0},\qquad
B=\frac{\rho_0\omega^2}{\gamma}\widehat\eta.
$$
This lower-fluid wave is outgoing; no additional incoming sound is included in defining the impedance seen by the upper fluid.

Put $K_s=Tk^2-m\omega^2$. The sheet's <force balance> gives $K_s\widehat\eta=-P+B$, hence $P=(\rho_0\omega^2/\gamma-K_s)\widehat\eta$. Its prescribed downward <velocity> amplitude is $V=-i\omega\widehat\eta$. Thus the <tensioned-sheet acoustic impedance> is
$$
\boxed{Z(k,\omega)=\frac{i\rho_0\omega}{\gamma}-\frac{i}{\omega}(Tk^2-m\omega^2)=Z_f+Z_s.}
$$
The first term is the lower fluid's <normal acoustic impedance>, and the second is the sheet's inertial and <elastic-sheet tension> response. For a real propagating angle, $Z_f=\rho_0c_0/\sin\theta$.

At fixed nonzero frequency and fixed wavenumber, $m\to\infty$ gives $|Z|\to\infty$, a zero-velocity, in-phase reflecting boundary. At fixed $k\ne0$, $T\to\infty$ gives the same reflection limit, but there is an important exception: \b[<elastic-sheet tension> does not resist the spatially uniform mode $k=0$]. At normal incidence, the impedance remains $\rho_0c_0+im\omega$ however large the <elastic-sheet tension> is. With $m=0$ this mode is transparent. By contrast, arbitrarily large <mass> resists even a spatially uniform oscillation. These fixed-frequency limits exclude a simultaneously tuned structural resonance.

If $m=T=0$, $Z=Z_f$ and $R=0$. The identical fluids are effectively joined across a massless, untensioned interface: <pressure> and <normal velocity> continue without reflection. This is the matched case, rather than the pressure-release case.